The set $S = \{1, 2, 3, \ldots , 49, 50\}$ contains the first $50$ positive integers.  After the multiples of 2 and the multiples of 3 are removed, how many integers remain in the set $S$?
Solution: The set $S$ contains $25$ multiples of 2 (that is, even numbers).  When these are removed, the set $S$ is left with only the odd integers from 1 to 49. At this point, there are $50-25=25$ integers in $S$. We still need to remove the multiples of 3 from $S$.

Since $S$ only contains odd integers after the multiples of 2 are removed,  we must remove the odd multiples of 3 between 1 and 49.  These are 3, 9, 15, 21, 27, 33, 39, 45, of which there are 8.  Therefore, the number of integers remaining in the set $S$ is $25 - 8 = \boxed{17}$.